BZOJ 2823 AHOI 2012 信号塔 凸包+最小圆覆盖

题目大意：给出平面上n个点，求最小圆覆盖。

思路：圆覆盖问题只与所有点中凸包上的点有关，因此先求一下凸包，然后数据范围骤减。大概是只剩下logn左右个点。这样就可以随便浪了。

先找所有三个点组成的圆，然后找两个点为直径所组成的圆。

还有就是三角形的外心公式，简直不是人推的，然后我就机制的百度了，结果如下：

不要模拟退火。。。

样例很坑，当你算出2.49 2.86的时候，不要认为你被卡精了，其实是你写挂了。。。

CODE：

#include <cmath>#include <cstdio>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#define MAX 1000010#define INF 1e15using namespace std;#define sqr(a) ((a) * (a)) struct Point{    double x,y,alpha;         Point(double _,double __):x(_),y(__) {}    Point() {}    bool operator <(const Point &a)const {        return alpha < a.alpha;    }    Point operator -(const Point &a)const {        return Point(x - a.x,y - a.y);    }    void Read() {        scanf("%lf%lf",&x,&y);    }    void Calc(const Point &p) {        alpha = atan2(y - p.y,x - p.x);    }}point[MAX]; int cnt;double min_x = INF,min_y = INF; Point stack[MAX];int top; double Cross(const Point &p,const Point &q){    return p.x * q.y - p.y * q.x;} double Calc(const Point &a,const Point &b){    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}double CalcX(const Point &p1,const Point &p2,const Point &p3){double up = sqr(p1.x) * (p2.y - p3.y) + sqr(p2.x) * (p3.y - p1.y) + sqr(p3.x) * (p1.y - p2.y);up -= (p1.y - p2.y) * (p2.y - p3.y) * (p3.y - p1.y);double down = p1.x * (p2.y - p3.y) + p2.x * (p3.y - p1.y) + p3.x * (p1.y - p2.y);return up / (2 * down);}double CalcY(const Point &p1,const Point &p2,const Point &p3){double up = -(sqr(p1.y) * (p2.x - p3.x) + sqr(p2.y) * (p3.x - p1.x) + sqr(p3.y) * (p1.x - p2.x));up += (p1.x - p2.x) * (p2.x - p3.x) * (p3.x - p1.x);double down = p1.x * (p2.y - p3.y) + p2.x * (p3.y - p1.y) + p3.x * (p1.y - p2.y);return up / (2 * down);}inline double Judge(double x,double y){double re = .0;for(int i = 0; i <= top; ++i)re = max(re,Calc(Point(x,y),stack[i]));return re;}int main(){    cin >> cnt;    for(int i = 1; i <= cnt; ++i)        point[i].Read();    int p;    for(int i = 1; i <= cnt; ++i)        if(point[i].y < min_y) {            min_y = point[i].y;            min_x = point[i].x;            p = i;        }        else if(point[i].y == min_y && point[i].x < min_x) {            min_x = point[i].x;            p = i;        }    for(int i = 1; i <= cnt; ++i)        if(i != p)            point[i].Calc(point[p]);    sort(point + 1,point + cnt + 1);    stack[top] = point[1];    stack[++top] = point[2];    stack[++top] = point[3];    for(int i = 4; i <= cnt; ++i) {        while(top >= 2 && Cross(stack[top] - stack[top - 1],point[i] - stack[top - 1]) <= 0)            --top;        stack[++top] = point[i];    }    double ans = INF;    double X,Y,x,y,r;    for(int i = 0; i <= top; ++i)    for(int j = i + 1; j <= top; ++j)    for(int k = j + 1; k <= top; ++k) {    x = CalcX(stack[i],stack[j],stack[k]);    y = CalcY(stack[i],stack[j],stack[k]);    r = Judge(x,y);    if(r < ans) {    ans = r;    X = x,Y = y;    }    }   for(int i = 0; i <= top; ++i)   for(int j = 0; j <= top; ++j) {r = Judge((stack[i].x + stack[j].x) / 2,(stack[i].y + stack[j].y) / 2);if(r < ans) {ans = r;X = (stack[i].x + stack[j].x) / 2;Y = (stack[i].y + stack[j].y) / 2;}}    printf("%.2lf %.2lf %.2lf\n",X,Y,ans);    return 0;}